University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.2
Symmetry factors: where the 1/4! fails to cancel
Every diagram you draw from here on carries a fraction in front, and a wrong one is the commonest slip in a first loop calculation. This lesson ends the guessing: the 1/(n!(4!)ⁿ) of the Dyson series is one over a group of relabellings, and what survives is one over the diagram's own symmetries, countable two independent ways.
Build the model
Connect the measurement to the mechanism.
At order n, φ⁴ perturbation theory gives every full contraction the same prefactor, (−iλ)ⁿ/(n!(4!)ⁿ), and n!(4!)ⁿ is no accident: it is the number of ways to rename the n dummy vertices and to permute the four identical fields at each. Those relabellings form a group G, and because they only rename integration variables and exchange equal operators, they carry every contraction to another with exactly the same integral. A Feynman diagram is one orbit of G: two contractions draw the same picture, external points fixed, exactly when a relabelling turns one into the other.
The orbit-stabiliser theorem does the rest. The orbit holds |G|/S contractions, S being the number of relabellings that leave one contraction unchanged, the diagram's automorphisms: flipping a self-loop, permuting parallel lines, exchanging vertices that no external point pins. Summed over its orbit the diagram keeps 1/S, derived: 2 for the tadpole, 8 for the figure-eight, 2 for the one-loop four-point bubble, 6 for the sunset, 1 for a tree whose ends are all labelled.
The argument stops where its two assumptions do. The four legs of a vertex must be one real field, so a vertex like ψ†ψφ, whose legs differ, keeps fewer symmetries; and every line end must be either a fixed external point or a summed vertex, so identical, integrated source legs put interchanges back: three on one vertex give 3! = 6.
- Simple definition
- A diagram's symmetry factor S is the number of ways to permute its vertices, and the line ends at each vertex, that give back exactly the same contraction, every line end still paired with the same partner and external points held fixed; the diagram enters the series with weight 1/S.
- Example
- The sunset joins x to z₁, z₁ to z₂ by three lines, and z₂ to y. Only the 3! orders of the three lines leave it unchanged, so S = 6: of the 945 contractions at order λ², 192 draw it, and 2!(4!)²/192 = 1152/192 = 6.
Derived: n! renames the dummy points z₁ … zₙ and each 4! permutes the four equal fields φ(z) at one vertex; neither changes a contraction's integral.
n the number of φ⁴ vertices; |G| is a pure number, 24 at n = 1 and 1152 at n = 2. The order-n term carries 1/|G|.
Derived: one diagram is one orbit of G, so the orbit-stabiliser theorem gives N/(n!(4!)ⁿ) = 1/S, the only factor the diagram keeps.
N the number of contractions that draw one diagram, S the number of relabellings that fix one of them; both pure numbers.
Derived: relabellings fixing every vertex can only permute parallel lines or flip and swap self-loops. Tadpole 2, figure-eight 2² × 2! = 8, sunset 3! = 6.
s: vertex permutations keeping the diagram, external points fixed; k: lines in a parallel bundle; ℓ: self-loops at a vertex.
Derived: 192 of the 945 order-λ² contractions draw it, and 1152/192 = 6. A left-right mirror would halve it wrongly, since x and y are never relabelled.
λ dimensionless, d⁴z in GeV⁻⁴, DF in GeV²: the term is GeV², like DF(x − y). The prefactor is (−iλ)²/6.
Derived: integrated sources are identical, so their 3! interchanges survive: N = 24, |G| = 3! × 4! = 144, S = 6, the λ/3! of the classical φ³ term (topic 3.09).
J a c-number source of mass dimension 3; the bracket has mass dimension 1, like φ, and so does the whole term.
Derived: each of the (4n + E − 1)!! Wick contractions lies in exactly one orbit. In zero dimensions, where every line is 1, it is a Gaussian moment.
Sum over every diagram, connected or not, with n vertices and E external points; n = 1, E = 2 gives 15/24 = 1/2 + 1/8.
The prefactor is one over a group
Work in ħ = c = 1 with the metric (+, −, −, −). Topic 3.01 wrote the order-n term of a φ⁴ correlation function as (1/n!)(−iλ/4!)ⁿ ∫d⁴z₁ … d⁴zₙ ⟨0|T φ(x₁) … φ(xE) φ(z₁)⁴ … φ(zₙ)⁴|0⟩, and Wick's theorem (topic 2.08) as a sum over (4n + E − 1)!! full contractions, each a product of DF's. Give each of the 4n vertex fields its own slot. Two kinds of relabelling leave the term alone. The zᵢ are dummy integration variables, so permuting their names, n! ways, changes nothing; and the four φ(z) at a vertex are one operator at one point, so permuting them, 4! ways per vertex, changes nothing either. Together they form a group G of order n!(4!)ⁿ, 24 at n = 1 and 1152 at n = 2, and G acts on contractions: relabel one and you get another whose integral is identical. The prefactor is exactly 1/|G|. The external points are not in G: they are fixed arguments of the correlation function, never summed, never renamed.
A diagram is an orbit, its symmetries a stabiliser
Call two contractions equivalent when an element of G maps one to the other. Equivalent contractions draw the same picture with the same external labels, and two contractions drawing the same picture differ only in how vertices and fields are named, so a Feynman diagram is exactly one orbit of G. The orbit-stabiliser theorem says the orbit holds N = |G|/S contractions, where S is the order of the stabiliser, the relabellings that send one contraction to itself. Every contraction in the orbit has the same integral, so the diagram's total is N/|G| = 1/S times (−iλ)ⁿ times one integral of DF's, derived. On the picture the stabiliser is the automorphism group: permutations of vertices and of line ends at each vertex that keep every connection, external points fixed. Where S = 1 the 4!'s and the n! cancel completely, which is why −iλ per vertex is the default. Where S ≥ 2, S different-looking relabellings of the fields return one and the same contraction, so the |G| relabellings produce only |G|/S distinct terms: there the 1/4! fails to cancel. Nothing here used the 4, so φᵏ theory with λ/k! has |G| = n!(k!)ⁿ and the same rule.
Counting automorphisms on the picture
Split an automorphism into what it does to the vertices and what it does at them. One that fixes every vertex can only reorder interchangeable line ends: k parallel lines joining the same two vertices can be permuted in k! ways, both ends moving together, and ℓ self-loops at one vertex can each be flipped and permuted among themselves, 2(ℓ) ℓ! ways. The rest permute vertices, and only vertices with no external point can move. So S = s × ∏k! × ∏2(ℓ) ℓ!, with s the allowed vertex permutations, derived. Tadpole: one self-loop, S = 2. Figure-eight: two self-loops at one vertex, S = 2² × 2! = 8. One-loop four-point bubble with x₁, x₂ on z₁ and x₃, x₄ on z₂: one double bundle, S = 2!, and no vertex swap, since it would carry x₁ to where x₃ is. Sunset: a triple bundle, S = 3! = 6, with z₁ and z₂ pinned by x and y. The vacuum melon, two vertices joined by four lines, has nothing pinned, so its swap counts: S = 2 × 4! = 48.
Checking each S by counting contractions
Counting contractions checks each automorphism count independently. Tadpole, n = 1: x picks one of 4 fields at z, y one of the 3 left, and the last two pair: N = 12, and 24/12 = 2. Figure-eight: the four fields pair in 3 ways, and 24/3 = 8; together 12 + 3 = 15 = 5!!, nothing missed. Bubble, n = 2: x₁ picks one of 8 fields, choosing its vertex; x₂ one of the 3 left there; x₃ one of the 4 at the other vertex and x₄ one of 3; the two fields left at each vertex join across in 2 ways: N = 8 × 3 × 4 × 3 × 2 = 576, and 1152/576 = 2. Sunset: x picks one of 8, y one of the 4 at the other vertex, and the three left at each vertex pair across in 3! ways: N = 8 × 4 × 6 = 192, and 1152/192 = 6. At order λ² the seven two-point diagrams, connected or not, hold 288 + 288 + 192 + 72 + 72 + 24 + 9 = 945 = 9!! contractions: the whole Wick count, each contraction in exactly one diagram.
Trees: labels remove symmetry, sources restore it
Every end of a φ⁴ tree is an external point, so each line, cut, splits the external points into two non-empty groups. The line ends at a vertex are told apart by the external points beyond them, so an automorphism fixing those points fixes every line and vertex: S = 1 for a tree whose ends are all labelled, shown. The order-λ four-point vertex has N = 4! = 24 = |G|, and the six-point tree at order λ², x₁, x₂, x₃ on one vertex and x₄, x₅, x₆ on the other, has N = 2 × 24 × 24 = 1152 = |G|. Add a source, ∫d⁴y J(y)φ(y), instead: k insertions of i∫Jφ arrive with 1/k! from their exponential, identical and integrated, and relabel like vertices. The tree with x on z and three source legs has |G| = 3! × 4! = 144 and N = 4 × 3! = 24, so S = 6, the 3! of the identical legs. Legs on different vertices multiply: x on z₁ with two sources, z₁ joined to z₂ with three, has S = 2! × 3! = 12, and N = 11520 = 138240/12 confirms it. Identical branches permute as wholes too: with x and one source on z₁, and z₁ joined to z₂ and z₃ that carry three sources each, swapping the two branches gives S = 2 × 3! × 3! = 72, and N = 5806080 = 3!(4!)³7!/72 confirms it. Topic 3.09 matches these to the classical φ³ term.
A zero-dimensional theory checks every S at once
Shrink spacetime to a point. The path integral becomes Z(λ) = ∫dφ e(−φ²/2 − λφ⁴/4!)/√(2π), the free two-point function becomes 1, and the Gaussian moments ⟨φ(2k)⟩₀ = (2k − 1)!! are Wick's theorem with every DF set to 1. Each vertex brings −λ in place of −iλ∫d⁴z, so each diagram is exactly (−λ)ⁿ/S; once the denominator has removed the vacuum bubbles (topic 3.03), the λⁿ coefficient of ⟨φ²⟩ is (−1)ⁿ times the sum of 1/S over connected two-point diagrams, derived. Expanding gives 1 − λ/2 + 2λ²/3 − 11λ³/8 + …: the −1/2 is the tadpole alone, and 2/3 = 1/4 + 1/4 + 1/6 is two tadpoles in series, a tadpole on a tadpole and the sunset. At λ = 0.010 the integral done numerically gives 0.9950653, the series to λ² 0.9950667 and to λ³ 0.9950653. No experiment measures a symmetry factor alone, but a wrong one is not harmless: the bubble's 1/2, in three channels, fixes topic 6.08's β(λ) = 3λ²/(16π²), quoted here, which would read 3λ²/(8π²) without it.
Change one variable at a time
Make the relationship visible.
Both sliders start at S = 1, the guess that every 4! cancels. Raise the tadpole's S until the dashed line meets the solid curve at λ = 0, then the sunset's: it lands on 6, not the 12 a left-right mirror would give, and S = 5 or 7 would need a fractional orbit.
TADPOLE ORBIT 4!/S (TRUE 12)24.0 contractions
SUNSET ORBIT 1152/S (TRUE 192)1152.0 contractions
λ COEFFICIENT (EXACT −0.500)-1.000
λ² COEFFICIENT (EXACT 0.667)1.500
Live interpretationTADPOLE ORBIT 4!/S (TRUE 12): 24.0 contractions. SUNSET ORBIT 1152/S (TRUE 192): 1152.0 contractions. λ COEFFICIENT (EXACT −0.500): −1.000. λ² COEFFICIENT (EXACT 0.667): 1.500
Catch the common trap
Explain before calculating.
In φ⁴ theory the two-loop sunset contribution to ⟨Ω|T φ(x)φ(y)|Ω⟩ is c (−iλ)² ∫d⁴z₁ d⁴z₂ DF(x − z₁) DF(z₁ − z₂)³ DF(z₂ − y). What is c?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe order-λ² vacuum melon joins two vertices z₁ and z₂ by four lines. Find its symmetry factor from its automorphisms and again by counting contractions, and give c in c(−iλ)² ∫d⁴z₁ d⁴z₂ DF(z₁ − z₂)⁴.
- Automorphisms: no external point pins either vertex, so the swap z₁ ↔ z₂ is allowed, s = 2; the four parallel lines permute, both ends together, in 4! = 24 ways. S = 2 × 24 = 48.
- Group: n = 2, so |G| = 2! × (4!)² = 2 × 576 = 1152.
- Contractions: every field at z₁ must pair with one at z₂. The first takes one of 4, the next one of 3, then 2, then 1: N = 4! = 24.
- Orbit-stabiliser: S = |G|/N = 1152/24 = 48, agreeing with the automorphism count, so c = 1/48.
- Sum-rule check: the other order-λ² vacuum diagrams are two figure-eights (S = 2 × 8 × 8 = 128, N = 9) and the dumbbell, a self-loop on each vertex and two lines between (S = 2 × 2! × 2 × 2 = 16, N = 72). 9 + 72 + 24 = 105 = 7!!, and 1/128 + 1/16 + 1/48 = 35/384 = 105/1152.
AnswerS = 48, so c = 1/48: 24 of the 105 = 7!! order-λ² vacuum contractions draw the melon.
MediumIn zero-dimensional φ⁴, ⟨φ²⟩ = ∫dφ φ² e(−φ²/2 − λφ⁴/24) ÷ ∫dφ e(−φ²/2 − λφ⁴/24). Expand both integrals to order λ² with Gaussian moments, find the λ and λ² coefficients of ⟨φ²⟩, match them to the symmetry factors of the connected two-point diagrams, and compare the series with the numerical value 0.9555766 at λ = 0.10.
- Gaussian moments with unit variance are ⟨φ(2k)⟩₀ = (2k − 1)!!: 1, 3, 15, 105, 945 for φ², φ⁴, φ⁶, φ⁸, φ¹⁰. Each counts pairings, Wick's theorem with every line equal to 1.
- Numerator over the free integral: 1 − (λ/24) × 15 + (λ²/(2 × 24²)) × 945 = 1 − 0.625λ + 0.8203λ². Denominator: 1 − (λ/24) × 3 + (λ²/1152) × 105 = 1 − 0.125λ + 0.09115λ².
- Divide, writing the ratio of 1 + aλ + bλ² to 1 + cλ + dλ² as 1 + (a − c)λ + (b − d − c(a − c))λ²: the λ coefficient is −0.625 + 0.125 = −0.500, the λ² coefficient 0.82031 − 0.09115 − 0.06250 = 0.66667.
- Diagrams: at order λ only the tadpole is connected, (−λ)/2, matching −1/2. At order λ², two tadpoles in series (S = 4), a tadpole on a tadpole (2! × 2 = 4) and the sunset (6) give 1/4 + 1/4 + 1/6 = 2/3. The division removed the vacuum pieces, as topic 3.03 proves in general.
- At λ = 0.10 the series gives 1 − 0.05 + 0.006667 = 0.95667, above the numerical 0.95558 by 0.00109; the next two terms, −11λ³/8 = −0.00138 and +34λ⁴/9 = +0.00038, close the gap to 0.00009.
Answer⟨φ²⟩ = 1 − λ/2 + 2λ²/3 + O(λ³), with −1/2 the tadpole's 1/S and 2/3 = 1/4 + 1/4 + 1/6; at λ = 0.10 the series gives 0.9567 against 0.9556.
HardAt order λ³ in ⟨Ω|T φ(x)φ(y)|Ω⟩, take the diagram with x and y both on z₁, one line from z₁ to z₂, one from z₁ to z₃, and three lines joining z₂ to z₃. Find S from its automorphisms and by counting contractions, write the term with its dimensions checked, and find its share of the order-λ³ zero-dimensional coefficient, −11/8.
- Automorphisms: z₁ is pinned by x and y, but z₂ and z₃ carry no external point and each joins z₁ once, so swapping them, links included, is allowed: s = 2. The triple bundle permutes in 3! = 6 ways; there are no self-loops. S = 2 × 6 = 12.
- Group: |G| = 3! × (4!)³ = 6 × 13824 = 82944.
- Contractions: x picks any of 12 fields, choosing z₁; y one of the 3 left there; of z₁'s last two fields the first joins one of the 8 on the other vertices and the second one of the 4 on the remaining vertex; the 3 + 3 fields left pair across in 3! = 6 ways. N = 12 × 3 × 8 × 4 × 6 = 6912, and S = 82944/6912 = 12.
- Term: (−iλ)³/12 = iλ³/12, times ∫d⁴z₁ d⁴z₂ d⁴z₃ DF(x − z₁) DF(y − z₁) DF(z₁ − z₂) DF(z₁ − z₃) DF(z₂ − z₃)³. Seven DF's give GeV¹⁴ and three d⁴z give GeV⁻¹², so the term is GeV², like DF(x − y). With I = 5 internal lines and V = 3, it has 5 − 3 + 1 = 3 loops.
- Zero dimensions: this diagram gives (−λ)³/12. The ten connected order-λ³ two-point diagrams have S = 4, 4, 8, 8, 8, 8, 8, 12, 12, 12, and 2/4 + 5/8 + 3/12 = 11/8, the size of the exact coefficient. This one supplies (1/12)/(11/8) = 6.1% of it.
AnswerS = 12 (2 for z₂ ↔ z₃, 3! for the triple line), confirmed by N = 6912 = 82944/12; the term is (iλ³/12) times its three-loop integral, about 6.1% of the λ³ coefficient in zero dimensions.